• The point estimate is in most cases a rude estimate. Rather, we may construct an interval for our parameter.

• Recall that the confidence interval \(\hat \Theta(y)\) (or region in the case of dimension of the parameter greater than \(1\)) for \(\theta\) with confidence level \(1-\alpha\) is a subset of the parameter space \(\Theta\), defined as function of the data \(y\) and such that \[Pr_\theta(\theta \in \hat \Theta(Y))= 1-\alpha \]

Let \(x=(x_1, \ldots, x_n)\) be the realizations of the r.v. \(X = (X_1, \ldots, X_n)\) where the \(X_i\), \(i=1, \ldots, n\), are independent and identically distributed according to \(\mathcal{N}(\mu, \sigma^2)\). To obtain a confidence interval for the mean \(\mu\) when \(\sigma^2\) is known, we leverage the pivotal quantity: \[ T(\mu)=\frac{\overline X - \mu}{\sigma/\sqrt{n}} \sim \mathcal{N}(0,1), \quad \forall \mu \in \mathbb{R}, \sigma^2 > 0.\]

Then, it follows that for \(0<\alpha <1\) \[\rm{Pr}(-z_{1-\alpha/2} \leq T(\mu) \leq z_{1-{\alpha/2}})=1-\alpha \] and a confidence interval of level \(1-\alpha\) for \(\mu\) is given by \[\text{CI}^{(1-\alpha)}_\mu=\bigg(\bar x - z_{1-\alpha/2} \frac{\sigma}{\sqrt{n}}, \bar x + z_{1-\alpha/2} \frac{\sigma}{\sqrt{n}}\bigg)\] where \(-z_{1-\alpha/2}=z_{\alpha/2}\) and \(z_{1-\alpha/2}\) represent the quantiles of order \(\alpha/2\) and \(1-\alpha/2\) of a standard normal distribution, respectively.

Also for interval estimation, Monte Carlo simulation is well suited to explore the procedure:

Also for interval estimation, Monte Carlo simulation is well suited to explore the procedure:

Let \(x=(x_1, \ldots, x_n)\) be the realization of the r.v. \(X = (X_1, \ldots, X_n)\) where the \(X_i\), \(i=1, \ldots, n\), are independent and identically distributed according to \(\mathcal{N}(\mu, \sigma^2)\). To obtain a confidence interval for the mean \(\mu\) when \(\sigma^2\) is unknown, and it is estimated by using the sample variance estimator \(S^2\), we leverage the pivotal quantity: \[ T(\mu)=\frac{\bar X - \mu}{S/\sqrt{n}} \sim t_{n-1}, \quad \forall \mu \in \mathbb{R}, \sigma^2>0 \]

Then, it follows that for \(0<\alpha <1\) \[\rm{Pr}(t_{n-1;\alpha/2} \leq T(\mu) \leq t_{n-1;1-{\alpha/2}})=1-\alpha \] and a confidence interval of level \(1-\alpha\) for \(\mu\) is given by \[\text{CI}^{1-\alpha}_\mu=\bigg(\bar x - t_{n-1; 1-\alpha/2} \frac{s}{\sqrt{n}}, \bar x + t_{n-1;1-\alpha/2} \frac{s}{\sqrt{n}}\bigg)\] where \(-t_{n-1;1-\alpha/2}=t_{n-1;\alpha/2}\) and \(t_{n-1;1-\alpha/2}\) represent the quantiles of order \(\alpha/2\) and \(1-\alpha/2\) of a t-student distribution with \(n-1\) degrees of freedom, respectively.

We have two populations and we collect data about samples from these populations. Let’s denote with \(y_{i1}, i = 1, \ldots, n_1\), a certain observed characteristic in the first sample and let \(y_{i2}\), \(i=1, \ldots, n_2\), the measurements on the same characteristic of the second sample. Here, we are following Section 4.5 of Agresti and Kateri’s book.

Let’s consider a quantitative variable. Goal: compare the population means \(\mu_1\) and \(\mu_2\)

• Parameter: \(\mu_1 - \mu_2\).
• Parameter estimate: \(\bar y_1 - \bar y_2\).
• Estimator: \(\bar Y_1 - \bar Y_2\).

Suppose to assume \(Y_{i1} \sim \mathcal{N}(\mu_1, \sigma^2_1)\), \(i=1, \ldots, n_1\) and \(Y_{i2} \sim \mathcal{N}(\mu_2, \sigma^2_2)\), \(i=1, \ldots, n_2\), with \(Y_{i1}\) and \(Y_{i2}\) independent. We will also assume \(\sigma^2_1 = \sigma^2_2 = \sigma^2\). Note that

• You cannot/don’t assume the normality: you could consider the CLT
• It does not hold \(\sigma^2_1=\sigma^2_2=\sigma^2\): there exists the Welch test (we can also verify by means of the hypothesis testing if this relation holds)
• In paired sample, under \(n_1=n_2\), \(Y_{i1}\) is not independent from \(Y_{i2}\): you can still work on the difference
  

Let \(\overline Y_1\) and \(\overline Y_2\) be the sample mean for the two groups, respectively. Let us identify the pivotal-quantity

You just know that \(\sigma^2_1=\sigma^2_2= \sigma^2\) but it is a parameter. You need to estimate it. Let’s define the pooled variance estimator

\[S^2= \frac{(n_1-1) S^2_1 + (n_2 - 1) S^2_2}{n_1 + n_2 - 2}\]

\[T = \frac{\overline Y_1 - \overline Y_2 - (\mu_1 - \mu_2) }{S\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}} \]

Now we know that \(T \sim t_{n_1+n_2-2}\) and the \((1-\alpha)%\) confidence interval for \(\mu_1-\mu_2\) is given by \[ CI ^{1-\alpha}_{\mu_1-\mu_2}= (\bar y_1 - \bar y_2 - t_{n_1+n_2-2; 1-\alpha/2} \times s \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}, \bar y_1 - \bar y_2 + t_{n_1+n_2-2; 1-\alpha/2} \times s \sqrt{\frac{1}{n_1} + \frac{1}{n_2}})\]

Let us consider this example. A study makes use of a cognitive behavioral therapy to treat a sample of teenage girls who suffered from anorexia. The study, like most such studies, also had a control group that received no treatment. Then researchers analyzed how the mean weight change compared for the treatment and control groups. The girls in the study were randomly assigned to the cognitive behavioral therapy (Group1) or to the control group (Group2).

Let \(\mu_1\) and \(\mu_2\) denote the mean weight gains (in pounds) for these groups.

Anor <- read.table("http://stat4ds.rwth-aachen.de/data/Anorexia.dat", 
                   header=TRUE)
# Get difference post-pre treatment for the group cb and c 
cogbehav <- Anor$after[Anor$therapy == "cb"] - Anor$before[Anor$therapy == "cb"]
control <- Anor$after[Anor$therapy == "c"] - Anor$before[Anor$therapy == "c"]


# Get the 95% CI via t.test function   
res <- t.test(cogbehav, control, var.equal = TRUE, conf.level = 0.95)
res$conf.int

[1] -0.680137  7.593930
attr(,"conf.level")
[1] 0.95

• The mean weight change for the cognitive behavioral therapy could be as much as 0.68 pounds lower or as much as 7.59 pounds higher than the mean weight change for the control group.
• The interval includes 0: it is plausible that the population means are identical (we can also see the results underlying the hypothesis test).
• The confidence interval is relatively wide: sample sizes are not large.

For two groups,let \(\pi_1\) and \(\pi_2\) denote population proportions for a binary variable of interest. Suppose to have independent samples \(Y_{i1}\), \(i=1, \ldots, n_1\) and \(Y_{i2}\), \(i=1, \ldots, n_2\) from \(Be(\pi_1)\) and \(Be(\pi_2)\). To obtain a confidence interval of level \(1-\alpha\) for \(\pi_1 - \pi_2\), we leverage the pivotal quantity

\[ Z = \frac{\Pi_1 - \Pi_2 - (\pi_1 - \pi_2)}{\sqrt{\hat \pi_1 (1 - \hat \pi_1)/n_1 + \hat \pi_2 (1 - \hat \pi_2)/n_2}}\] which is approximately distributed as \(\mathcal{N}(0,1)\). Above, we denoted with \(\Pi_1\) and \(\Pi_2\) the sample proportion estimators. So the realisations of a confidence interval is given by

\[CI^{\pi_1 - \pi_2}_{1-\alpha} = (\hat \pi_1 - \hat \pi_2 \pm z_{1-\alpha/2} \sqrt{\hat \pi_1 (1 - \hat \pi_1)/n_1 + \hat \pi_2 (1 - \hat \pi_2)/n_2}) \]

A study used patients at six U.S. hospitals who were to receive coronary artery bypass graft surgery. The patients were randomly assigned to two groups. For one group, Christian volunteers were instructed to pray for a successful surgery with a quick, healthy recovery and no complications. The praying started the night before surgery and continued for two weeks. The other group did not have volunteers praying for them. The response was whether medical complications occurred within \(30\) days of the surgery

Let \(\pi_1\) denote the probability of complications for those patients who had a prayer group. Let \(\pi_2\) denote the probability of complications for the patients not having a prayer group.